An antifreeze solution is prepared from 222.6 g of ethylene glycol (C 2 H 6 O 2 ) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL –1 , then what shall be the molarity of the solution?

Question

Philoid · Accepted Answer

Given:

Mass of ethylene glycol (C₂H₆O₂) = 222.6 g

Mass of water = 200 g

Density, d = 1.072 g/ml

To find: Molality and Molarity of solution

Formula:

Molality

Molarity, M_o

Density, d

Solution:

Calculation of Molality:

⇒ Molecular Mass of ethylene glycol (C₂H₆O₂) = [12 × 2] + [6 × 1] + [16 × 2]

= 24 + 6 + 32

= 62 g

⇒ Number of moles of ethylene glycol (C₂H₆O₂) = [222.6/62]

= 3.59 moles

⇒ Mass of water = 200 g

⇒ Molality

= 17.95 m

Calculation of Molarity:

⇒ Total mass of solution = [222.6 + 200]

= 422.6 g

From density formula we can find out the volume required.

⇒ Volume of solution, V = [422.6 /1.072]

= 394.216 ml

⇒ Molarity, M_o

⇒ Molarity = 9.1067

⇒ Molarity ≈ 9.11 M

Therefore the Molality and Molarity of the solution is as follows:

Molality = 17.95 m

Molarity = 9.11 M

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^–1, then what shall be the molarity of the solution?